Chemistry II
posted by Cynthia .
How many grams of gasoline would you need to burn if you started with a 275 gram piece of ice that started at 25 OC and by the end of the experiment had 100 grams of steam and the rest liquid water?

Chemistry II 
DrBob222
We prefer you stick to the same screen name.
This is a many step problem. How much have you done and exactly what do you not understand. 
Chemistry II 
Jessica
CpxmassxTfTi
(2.06)(275)(0(25))=14162.5
333J/100x14162.5 = 47161.125
47161.125+14162.5= 61323.625
61323.625(1gas/48,000)= answer: 1.277g of steam
Thank you to the way for helping chemistry isn't my best subject so I apreciate the time y'all put in to help tutor 
Chemistry II 
DrBob222
CpxmassxTfTi
(2.06)(275)(0(25))=14162.5 This step is perfect. Good work!
333J/100x14162.5 = 47161.125 I don't understand this step. You should be melting all 275 g of the ice so it should be 333 J/g x 275 g = ? J for the melting step.
Now you want to raise the temperature of the 275 g water from zero C to 100 C which is 275g x 4.184 J/g x (1000) = ?
Then you want to vaporize 100 g of the 275 g H2O you have to steam which will leave you with 175 g of liquid water.
That is 100 g H2O x heat vaporization which I think is about 2260 J/g (but check me out on that for sometimes I don't remember all of these constants) = ? J.
All all of those values together to find total q needed.
Then look up the heat of combustion of gasoline and we can talk about that step when you get there.
47161.125+14162.5= 61323.625
61323.625(1gas/48,000)= answer: 1.277g of steam
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